Page 135 - A Course in Linear Algebra with Applications

5.1: Existence of a Basis 119

Now X can be written in the form
4
/ - / 3 \ /-4/3N
- 1 / 3 2/3
X = c + d 0
\ J/ V i/

where c and d are arbitrary scalars. Hence the null space of
A is generated by the vectors

-4/3 \
(
2/3
X, = Xo
0
\ 1 /

Notice that these vectors are obtained from the general solu-
tion X by putting c = 1, d = 0, and then c = 0, d = 1. Now
Xi and Xi are linearly independent. Indeed, if we assume
that some linear combination of them is zero, then, because
of the configuration of 0's and l's, the scalars are forced to be
be zero. It follows that X\ and X 2 form a basis of the null
space of A, which therefore has dimension equal to 2.
It should be clear to the reader that this example de-
scribes a general method for finding a basis, and hence the di-
mension, of the null space of an arbitrary mxn matrix A. The
procedure goes as follows. Using elementary row operations,
put A in reduced row echelon form, with say r pivots. Then
the general solution of the linear system AX = 0 will con-
tain n — r arbitrary scalars, say ci, C2,..., c n _ r . The method
of solving linear systems by elementary row operations shows
that the general solution can be written in the form

X = c\Xx + C2X2 + • • • + c n- rX n— r

where Xi,..., X n- r are particular solutions. In fact the solu-
tion Xi arises from X when we put c; = 1 and all other Cj's